表达式求值

Posted 2022-12-05 夏芷雨涵梦

题目： 使用栈来对表达式进行求值。对具有多个操作符和括号的复合表达式进行求值，假设操作数都是整数，运算符只包括+、-、×、/四种。

时间：2019.04.05

代码：

import java.util.Stack;

public class EvaluateExpression 
    public static void main(String []args)
        if(args.length!=1)
            System.out.println("Usage: java EvaluateExpression \\"expression\\"");
            System.exit(1);
        
        //args[0]就是用命令行运行java程序时传入的第一个参数
        try
            System.out.println(evaluateExpression(args[0]));
        catch(Exception e)
            System.out.println("Wrong expression: "+args[0]);
        
    
    public static int evaluateExpression(String expression)
        Stack<Integer> operandStack=new Stack<>();
        Stack<Character>operatorStack=new Stack<>();
        expression=insertBlanks(expression);
        String []tokens=expression.split(" ");
        for(String token:tokens)
            if(token.length()==0)
                continue;
            else if(token.charAt(0)=='+'||token.charAt(0)=='-')
                while(!operatorStack.isEmpty()&&(operatorStack.peek()=='+'||operatorStack.peek()=='-'
                ||operatorStack.peek()=='*'||operatorStack.peek()=='/'))
                    processAnOperator(operandStack,operatorStack);
                operatorStack.push(token.charAt(0));
            
            else if(token.charAt(0)=='*'||token.charAt(0)=='/')
                while(!operatorStack.isEmpty()&&(operatorStack.peek()=='*'||operatorStack.peek()=='/'))
                    processAnOperator(operandStack,operatorStack);
                operatorStack.push(token.charAt(0));
            
            //trim()去掉两端多余的空格
            else if(token.trim().charAt(0)=='(')
                operatorStack.push('(');
            
            else if(token.trim().charAt(0)==')')
                while (operatorStack.peek()!='(')
                    processAnOperator(operandStack,operatorStack);
                operatorStack.pop();
            
            else
                operandStack.push(new Integer(token));
        
        while(!operatorStack.isEmpty())
            processAnOperator(operandStack,operatorStack);
        
        return operandStack.pop();
    
    public static void processAnOperator(Stack<Integer> operandStack,Stack<Character> operatorStack)
        char op=operatorStack.pop();
        int op1=operandStack.pop();
        int op2=operandStack.pop();
        if(op=='+')
            operandStack.push(op2+op1);
        else if(op=='-')
            operandStack.push(op2-op1);
        else if(op=='*')
            operandStack.push(op2*op1);
        else if(op=='/')
            operandStack.push(op2/op1);
    
    //用于保证操作数、操作符和括号至少被一个空格分隔
    public static String insertBlanks(String s)
        String result="";
        for (int i=0;i<s.length();i++)
            if(s.charAt(i)=='('||s.charAt(i)==')'||s.charAt(i)=='+'||s.charAt(i)=='-'
            ||s.charAt(i)=='*'||s.charAt(i)=='/')
                result+=" "+s.charAt(i)+" ";//用空格分隔开
            else
                result+=s.charAt(i);
        
        return result;
    

该问题使用两个栈来解决问题，命名为operandStack和operatorStack，分别用于存放操作数和操作符。操作数和操作符在被处理前被压入栈中。当一个操作符被处理时，它从operatorStack中弹出，并应用于operandStaack中的前两个操作数，结果被压回operandStack中。